Help for a problem with inscribed triangles

If we have a triangle $ABC$ with $AB = 3\sqrt 7$, $AC = 3$, $\angle{ACB} = \pi/3$, $CL$ is the bisector of angle $ACB$, $CL$ lies on line $CD$ and $D$ is a point of the circumcircle of triangle $ABC$, what's the length of $CD$? Here I attach my solution. The problem is that I get $CD = 5\sqrt 3$ while in my text book the solution is given as $4\sqrt 3$ and I really cannot understand where I'm doing wrong. Could somebody help me out?

Your problem is in stating that $\cos \beta = \frac {\sqrt {7}} {14}$. This is incorrect.

Although it is true that $\cos^2 \beta = \frac {7} {196}$, there are two possible values for $\cos \beta$. If you were to make a scale drawing, you would find that AB is over twice as long as Ac, making $\beta$ an obtuse angle.

Thus $\cos \beta = -\frac {\sqrt {7}} {14}$

• A nice catch. $BC^2\gt AB^2+CA^2$ shows that $\beta$ is an obtuse angle. May 18, 2015 at 23:17
• Thank you both for the help! I was so frustrated over that problem, I triplechecked things but couldn't see that error. Great job :) May 18, 2015 at 23:40

As you wrote the radius $R$ of the circumcircle of $\triangle{ABC}$ is $\sqrt{21}$.

Since the radius of the circumcircle of $\triangle{ADC}$ is also $R=\sqrt{21}$, by the law of sines, $$2R=\frac{AD}{\sin{\angle{ACD}}}\Rightarrow AD=\sqrt{21}.$$ Then, applying the law of cosines to $\triangle{ACD}$ gives you $$21=3^2+CD^2-2\cdot 3\cdot CD\cdot \cos{\angle{ACD}}\Rightarrow CD=4\sqrt 3.$$

• Thank you for your input. I was looking for someone to point out a mistake in my calculations though and not give alternative solution. May 18, 2015 at 23:02
• Thanks for proving $4\sqrt{3}$ is the right answer. I'll continue looking for a bug. May 18, 2015 at 23:02
• I hope I've given you what you were looking for.
– tomi
May 18, 2015 at 23:05